The Golden Ratio

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{{Image Description Ready
{{Image Description Ready
|ImageName=The Golden Ratio
|ImageName=The Golden Ratio
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|Image=180px-Pentagram-phi.svg.png
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|Image=Goldenratioapplet1.jpg
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|ImageIntro=The pentagram at the right is designed using two isosceles triangles that exhibit the golden ratio. One triangle has a base of length 1 and legs of length <math>\varphi</math>, while the other triangle has a base of length <math>\varphi</math> and legs of length 1.
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|ImageIntro=The '''golden number,''' often denoted by lowercase Greek letter "phi", is <br /><math>{\varphi}=\frac{1 + \sqrt{5}}{2} = 1.61803399...</math>. <br />
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The golden number, often referred to as ''phi'' is numerically equal to <math>\frac{1 + \sqrt{5}}{2} \approx 1.61803399 \dots =\varphi</math>.
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The term '''golden ratio''' refers to any ratio which has the value phi. The image to the right illustrates dividing and subdividing a rectangle into the golden ratio. The result is [[Field:Fractals|fractal-like]]. This page explores real world applications for the golden ratio, common misconceptions about the golden ratio, and multiple derivations of the golden number.
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The term ''golden ratio'' refers to the ratio <math>\varphi</math> : 1.
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|ImageDescElem=[[Image:Monalisa01.jpg|Does the Mona Lisa exhibit the golden ratio?|thumb|400px|right]]The golden number, approximately 1.618, is called golden because many geometric figures involving this ratio are often said to possess special beauty. Be that true or not, the ratio has many beautiful and surprising mathematical properties. The Greeks were aware of the golden ratio, but did not consider it particularly significant with respect to aesthetics. It was not called the "divine" proportion until the 15th century, and was not called "golden" ratio until the 18th century. <ref>[http://en.wikipedia.org/wiki/Golden_ratio "Golden ratio"], Retrieved on 20 June 2012.</ref>
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<br /> [[Image:Finalpyramid1.jpg|Markowsky has determined the above dimensions to be incorrect.|thumb|400px|left]]
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Since then, it has been claimed that the golden ratio is the most aesthetically pleasing ratio, and claimed that this ratio has appeared in architecture and art throughout history. Among the most common such claims are that the Parthenon and Leonardo Da Vinci's Mona Lisa use the golden ratio. Even more esoteric claims propose that the golden ratio can be found in the human facial structure, the behavior of the stock market, and the Great Pyramids. <br />
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However, such claims have been criticized in scholarly journals as wishful thinking or sloppy mathematical analysis. Additionally, there is no solid evidence that supports the claim that the golden rectangle is the most aesthetically pleasing rectangle.<ref>[http://www.math.nus.edu.sg/aslaksen/teaching/maa/markowsky.pdf "Misconceptions about the Golden Ratio"], Retrieved on 24 June 2012.</ref>
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This page explores real world applications for the golden ratio, common misconceptions about the golden ratio, and multiple derivations of the golden number.
 
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|ImageDescElem=The golden ratio, approximately 1.618, is called golden because many geometric figures involving this ratio are often said to possess special beauty. Be that true or not, the ratio has many beautiful and surprising mathematical properties. The Greeks were aware of the golden ratio, but did not consider it particularly significant with respect to aesthetics. It was not called the "divine" proportion until the 15th century, and was not called "golden" ratio until the 18th century. Since then, it has been claimed that the golden ratio is the most aesthetically pleasing ratio, and claimed that this ratio has appeared in architecture and art throughout history. Among the most common such claims are that the Parthenon and Leonardo Da Vinci's Mona Lisa uses the golden ratio. Even more esoteric claims propose that the golden ratio can be found in the human facial structure, the behavior of the stock market, and the Great Pyramids. However, such claims have been criticized in scholarly journals (see references at the end of the page) as wishful thinking or sloppy mathematical analysis. Additionally, there is no solid evidence that supports the claim that the golden rectangle is the most aesthetically pleasing rectangle.
 
===Misconceptions about the Golden Ratio===
===Misconceptions about the Golden Ratio===
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In his paper, ''Misconceptions about the Golden Ratio,'' George Markowsky investigates many claims about the golden ratio appearing in man-made objects and in nature. Specifically, he claims that the golden ratio does not appear in the Parthenon or the Great Pyramids, two of the more common beliefs. He also disputes the belief that the human body exhibits the golden ratio. To read more, [http://www.math.nus.edu.sg/aslaksen/teaching/maa/markowsky.pdf click here!]
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Many rumors and misconceptions surround the golden ratio. There have been many claims that the golden ratio appears in art and architecture. In reality, many of these claims involve warped images and large margins of error. One claim is that the Great Pyramids exhibit the golden ratio in their construction. This belief is illustrated below.
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In his paper, ''Misconceptions about the Golden Ratio,'' George Markowsky disputes this claim, arguing that the dimensions assumed in the picture are not anywhere close to being correct. Another belief is that a series of [[The Golden Ratio#Jump2|golden rectangles]] appears in the ''Mona Lisa''.
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However, the placing of the golden rectangles seems arbitrary. Markowsky also disputes the belief that the human body exhibits the golden ratio. To read more, [http://www.math.nus.edu.sg/aslaksen/teaching/maa/markowsky.pdf click here!]
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:
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====''What do you think?''====
====''What do you think?''====
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[[Image:Golden ratio parthenon.jpg|300px]]
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George Markowsky argues that, like the ''Mona Lisa,'' the Parthenon does not exhibit a series of golden rectangles (discussed below). Do you think the Parthenon was designed with the golden ratio in mind or is the image below simply a stretch of the imagination?
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<ref>[http://lotsasplainin.blogspot.com/2008/01/wednesday-math-vol-8-phi-golden-ratio.html "Parthenon"], Retrieved on 16 May 2012.</ref>
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:[[Image:Golden ratio parthenon.jpg|300px]]<ref>[http://lotsasplainin.blogspot.com/2008/01/wednesday-math-vol-8-phi-golden-ratio.html "Parthenon"], Retrieved on 16 May 2012.</ref>
==A Geometric Representation==
==A Geometric Representation==
===The Golden Ratio in a Line Segment===
===The Golden Ratio in a Line Segment===
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[[Image:Golden_segment.jpg|400px]][[Image:Goldenratiolabeled1.jpg]]
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[[Image:Golden_segment.jpg|400px]][[Image:Animation2.gif]]
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The golden ratio can be defined using a line segment divided into two sections, of lengths a and b, respectively. If a and b are appropriately chosen, the ratio of a to b is the same as the ratio of a + b to a and both ratios are equal to <math>\varphi</math>. The value of this ratio turns out not to depend on the particular values of a and b, as long as they satisfy the proportion. The line segment above exhibits the golden proportions.
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The golden number can be defined using a line segment divided into two sections of lengths ''a'' and ''b''. If ''a'' and ''b'' are appropriately chosen, the ratio of ''a'' to ''b'' is the same as the ratio of ''a'' + ''b'' to ''a'' and both ratios are equal to <math>\varphi</math>. The line segment above (left) exhibits the golden proportion. The line segments above (right) are also examples of the golden ratio. In each case,
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The golden rectangle is made up of line segments exhibiting the golden proportion. Remarkably, when a square is cut off of the golden rectangle, the remaining rectangle also exhibits the golden proportions. This continuing pattern is visible in the golden rectangle above.
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<math>\frac{{\color{Red}\mathrm{red}}+\color{Blue}\mathrm{blue}}{{\color{Blue}\mathrm{blue}} }= \frac{{\color{Blue}\mathrm{blue}} }{{\color{Red}\mathrm{red}} }= \varphi . </math>
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<div id="Jump2"></div>
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===The Golden Rectangle===
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A '''golden rectangle''' is any rectangle where the ratio between the sides is equal to phi. When the sides lengths are proportioned in the golden ratio, the rectangle is said to possess the '''golden proportions.''' A golden rectangle has sides of length <math>\varphi \times r</math> and <math>1 \times r</math> where <math>r</math> can be any constant. Remarkably, when a square with side length equal to the shorter side of the rectangle is cut off from one side of the golden rectangle, the remaining rectangle also exhibits the golden proportions. This continuing pattern is visible in the golden rectangle below.
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:[[Image:Coloredfinalrectangle1.jpg]]
===Triangles===
===Triangles===
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[[Image:Final2triangles.jpg|500px]][[Image:Pentagon_final.jpg|300px]]
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[[Image:1byrrectangle1.jpg|500px]][[Image:Pentagon_final.jpg|300px]]
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The golden ratio <math>\varphi</math> is used to construct the golden triangle, an isoceles triangle that has legs of length <math>\varphi</math> and base length of 1. It is above and to the left. Similarly, the golden gnomon has base <math>{\varphi}</math> and legs of length 1. It is shown above and to the right. These triangles can be used to form <balloon title="A pentagram is a five pointed star made with 5 straight strokes">pentagrams</balloon> and <balloon title="A pentacle is a five pointed star inscribed in a circle.">pentacles</balloon>.
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The golden number, <math>\varphi</math>, is used to construct the '''golden triangle,''' an isoceles triangle that has legs of length <math>\varphi \times r</math> and base length of <math>1 \times r</math> where <math>r</math> can be any constant. It is above and to the left. Similarly, the '''golden gnomon''' has base <math>\varphi \times r</math> and legs of length <math>1 \times r</math>. It is shown above and to the right. These triangles can be used to form regular pentagons (pictured above) and <balloon title="A pentagram is a five pointed star made with 5 straight strokes">pentagrams.</balloon>
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The pentacle below, generated by the golden triangle and the golden gnomon, has many side lengths proportioned in the golden ratio.
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The pentgram below, generated by the golden triangle and the golden gnomon, has many side lengths proportioned in the golden ratio.
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:[[Image:Uprightpentacle2.jpg]]
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:[[Image:Star1.jpg]]
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:<math>\frac{\mathrm{blue} }{\mathrm{red} } = \frac{\mathrm{red} }{\mathrm{green} } = \frac{\mathrm{green} }{\mathrm{pink} } = \varphi . </math>
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:::<math>\frac{{\color{SkyBlue}\mathrm{blue}} }{{\color{Red}\mathrm{red}} } = \frac{{\color{Red}\mathrm{red}} }{{\color{Green}\mathrm{green}} } = \frac{{\color{Green}\mathrm{green}} }{{\color{Magenta}\mathrm{pink}} } = \varphi . </math>
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These triangles can be used to form [[Field:Fractals| fractals]] and are one of the only ways to tile a plane using pentagonal symmetry. Two fractal examples are shown below.
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These triangles can be used to form [[Field:Fractals| fractals]] and are one of the only ways to tile a plane using '''pentagonal symmetry'''. Pentagonal symmetry is best explained through example. Below, we have two fractal examples of pentagonal symmetry. Images that exhibit pentagonal symmetry have five symmetry axes. This means that we can draw five lines from the image's center, and all resulting divisions are identical.
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[[Image:Penrose-4.jpg|250px]] [[Image:Penrose-21.jpg|250px]]
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:[[Image:Penta1.jpg|400px]]
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|ImageDesc==Mathematical Representations of the Golden Ratio=
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:[[Image:Pent111.jpg|400px]]
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|ImageDesc==An Algebraic Derivation of Phi=
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{{SwitchPreview|ShowMessage=Click to expand|hideMessage=Click to hide|PreviewText=How can we derive the value of <math>\varphi</math> from its characteristics as a ratio? We may algebraically solve for the ratio (<math>\varphi</math>) by observing that ratio satisfies the following property by definition:
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==An Algebraic Representation==
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{{SwitchPreview|ShowMessage=Click to expand|HideMessage=Click to hide|PreviewText=We may algebraically solve for the ratio (<math> \varphi </math>) by observing that ratio satisfies the following property by definition:
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:<math>\frac{b}{a} = \frac{a+b}{b} = \varphi</math>|FullText=
:<math>\frac{b}{a} = \frac{a+b}{b} = \varphi</math>|FullText=
Let <math> r </math> denote the ratio :
Let <math> r </math> denote the ratio :
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So
So
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:<math>r=\frac{a+b}{a}=1+\frac{b}{a} =1+\cfrac{1}{a/b}=1+\frac{1}{r}</math>.
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:<math>r=\frac{a+b}{a}=1+\frac{b}{a}</math> which can be rewritten as
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:<math>1+\cfrac{1}{a/b}=1+\frac{1}{r}</math> thus,
:<math>r=1+\frac{1}{r}</math>
:<math>r=1+\frac{1}{r}</math>
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</span>, we get <math>r = \frac{1 \pm \sqrt{5}} {2}</math>.
</span>, we get <math>r = \frac{1 \pm \sqrt{5}} {2}</math>.
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Because the ratio has to be a positive value,
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The ratio must be positive because we can not have negative line segments or side lengths. Because the ratio has to be a positive value,
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:<math>r=\frac{1 + \sqrt{5}}{2} \approx 1.61803399 \dots =\varphi</math>.
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:<math>r=\frac{1 + \sqrt{5}}{2} = 1.61803399... =\varphi</math>.
|NumChars=500}}
|NumChars=500}}
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==Continued Fraction Representation and [[Fibonacci sequence|Fibonacci Sequences]]==
==Continued Fraction Representation and [[Fibonacci sequence|Fibonacci Sequences]]==
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The golden ratio can also be written as what is called a '''continued fraction''' by using <balloon title="Recursion is the method of substituting an equation into itself">recursion</balloon>.
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The golden ratio can also be written as what is called a '''continued fraction,'''a fraction of infinite length whose denominator is a quantity plus a fraction, which latter fraction has a similar denominator, and so on. This is done by using <balloon title="Recursion is the method of substituting an equation into itself">recursion</balloon>.
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{{SwitchPreview|ShowMessage=Click to expand|HideMessage=Click to hide|PreviewText= |FullText=We have already solved for <math> \varphi </math> using the following equation:
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{{SwitchPreview|ShowMessage=Click to expand|hideMessage=Click to hide|PreviewText= |FullText=We have already solved for <math>\varphi</math> using the following equation:
<math>{\varphi}^2-{\varphi}-1=0</math>.
<math>{\varphi}^2-{\varphi}-1=0</math>.
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<math>\varphi -1= \cfrac{1}{\varphi }</math>.
<math>\varphi -1= \cfrac{1}{\varphi }</math>.
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Solving for <math> \varphi </math> gives
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Adding 1 to both sides gives
<math>\varphi =1+ \cfrac{1}{\varphi }</math>.
<math>\varphi =1+ \cfrac{1}{\varphi }</math>.
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Now use recursion and substitute in the entire right side of the equation for <math> \varphi </math> in the bottom of the fraction.
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Substitute in the entire right side of the equation for <math> \varphi </math> in the bottom of the fraction.
<math>\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{\varphi } }</math>
<math>\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{\varphi } }</math>
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<math>\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\cdots}}}</math>
<math>\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\cdots}}}</math>
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This last infinite form is a continued fraction
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Continuing this substitution forever, the last infinite form is a continued fraction
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If we evaluate truncations of the continued fraction by evaluating only part of the continued fraction (the finite displays above it), replacing <math>\varphi</math> by 1, we produce the ratios between consecutive terms in the [[Fibonacci sequence]].
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If we evaluate truncations of the continued fraction by evaluating only part of the continued fraction, replacing <math>\varphi</math> with 1, we produce the ratios between consecutive terms in the [[Fibonacci sequence]].
<math>\varphi \approx 1 + \cfrac{1}{1} = 2</math>
<math>\varphi \approx 1 + \cfrac{1}{1} = 2</math>
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<math>\varphi \approx 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1+\cfrac{1}{1}}}} = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{2}}} =1 + \cfrac{1}{1 + \cfrac{2}{3}} = 8/5</math>
<math>\varphi \approx 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1+\cfrac{1}{1}}}} = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{2}}} =1 + \cfrac{1}{1 + \cfrac{2}{3}} = 8/5</math>
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Thus we discover that the golden ratio is approximated in the [[Fibonacci sequence]].
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Thus we discover that the golden ratio is approximated in the Fibonacci sequence.
<math>1,1,2,3,5,8,13,21,34,55,89,144...\,</math>
<math>1,1,2,3,5,8,13,21,34,55,89,144...\,</math>
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<math>\varphi = 1.61803399...\,</math>
<math>\varphi = 1.61803399...\,</math>
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As you go farther along in the [[Fibonacci sequence]], the ratio between the consecutive terms approaches the golden ratio. Many real world applications of the golden ratio are related to the [[Fibonacci sequence]]. For more real-world applications of the golden ratio [[Fibonacci Numbers|click here!]]
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As you go farther along in the Fibonacci sequence, the ratio between the consecutive terms approaches the golden ratio. Many real-world applications of the golden ratio are related to the Fibonacci sequence. For more real-world applications of the golden ratio [[Fibonacci Numbers|click here!]]
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In fact, we can prove this relationship using mathematical [[Induction]].
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In fact, we can prove that the ratio between terms in the Fibonacci sequence approaches the golden ratio by using mathematical [[Induction]].
{{Switch|link1=Click to show proof|link2=Click to hide proof|1= |2=
{{Switch|link1=Click to show proof|link2=Click to hide proof|1= |2=
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we only need to show that each of the terms in the continued fraction is the ratio of Fibonacci numbers as shown above.
we only need to show that each of the terms in the continued fraction is the ratio of Fibonacci numbers as shown above.
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First, let <math> x_1=1</math>, <math> x_2=1+\frac{1}{1}=1+\frac{1}{x_1} </math>, <math> x_3= 1+\frac{1}{1+\frac{1}{1}}=1+\frac{1}{x_2} </math> and so on so that <math> x_n=1+\frac{1}{x_{n-1}} </math>.
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First, let
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:<math> x_1=1</math>,
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:<math> x_2=1+\frac{1}{1}=1+\frac{1}{x_1} </math>,
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:<math> x_3= 1+\frac{1}{1+\frac{1}{1}}=1+\frac{1}{x_2} </math> and so on so that
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:<math> x_n=1+\frac{1}{x_{n-1}} </math>.
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These are just the same truncated terms as listed above. Let's also denote the terms of the [[Fibonacci sequence]] as <math> f_n=f_{n-1}+f_{n-2} </math> where <math>f_1=1</math>,<math>f_2=1</math>, and so <math>f_3=1+1=2</math>, <math> f_4=1+2=3 </math> and so on.
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These are just the same truncated terms as listed above. Let's also denote the terms of the Fibonacci sequence as
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:<math> s_n=s_{n-1}+s_{n-2} </math> where <math>s_1=1</math>,<math>s_2=1</math>,<math>s_3=2</math>,<math>s_4=3</math> etc.
<br>
<br>
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We want to show that <math> x_n=\frac{f_{n+1}}{f_n} </math> for all n.
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We want to show that
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:<math> x_n=\frac{s_{n+1}}{s_n} </math> for all n.
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First, we establish our [[Induction|base case]]. We see that <math> x_1=1=\frac{1}{1}=\frac{f_2}{f_1} </math>, and so the relationship holds for the base case.
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First, we establish our [[Induction|base case]]. We see that
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:<math> x_1=1=\frac{1}{1}=\frac{s_2}{s_1} </math>, and so the relationship holds for the base case.
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Now we assume that <math> x_k=\frac{f_{k+1}}{f_{k}} </math> for some <math> 1 \leq k < n </math> (This step is the [[Induction|inductive hypothesis]]). We will show that this implies that <math> x_{k+1}=\frac{f_{(k+1)+1}}{f_{k+1}}=\frac{f_{k+2}}{f_{k+1}} </math>.
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Now we assume that
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:<math> x_k=\frac{s_{k+1}}{s_{k}} </math> for some <math> 1 \leq k < n </math> (This step is the [[Induction|inductive hypothesis]]). We will show that this implies that
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:<math> x_{k+1}=\frac{s_{(k+1)+1}}{s_{k+1}}=\frac{s_{k+2}}{s_{k+1}} </math>.
<br><br>
<br><br>
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By our definition of <math>x_n</math>, we have
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By our assumptions about ''x<sub>1</sub>'',''x<sub>2</sub>''...''x<sub>n</sub>'', we have
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<math> x_{k+1}=1+\frac{1}{x_k} </math>.
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:<math> x_{k+1}=1+\frac{1}{x_k} </math>.
By our inductive hypothesis, this is equivalent to
By our inductive hypothesis, this is equivalent to
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<math>x_{k+1}=1+\frac{1}{\frac{f_{k+1}}{f_{k}}}</math>.
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:<math>x_{k+1}=1+\frac{1}{\frac{s_{k+1}}{s_{k}}}</math>.
Now we only need to complete some simple algebra to see
Now we only need to complete some simple algebra to see
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<math> x_{k+1}=1+\frac{f_k}{f_{k+1}} </math>
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:<math> x_{k+1}=1+\frac{s_k}{s_{k+1}} </math>
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<math> x_{k+1}=\frac{f_{k+1}+f_k}{f_{k+1}} </math>
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:<math> x_{k+1}=\frac{s_{k+1}+s_k}{s_{k+1}} </math>
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Noting the definition of <math>f_n=f_{n-1}+f_{n-2}</math>, we see that we have
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Noting the definition of <math>s_n=s_{n-1}+s_{n-2}</math>, we see that we have
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<math> x_{k+1}=\frac{f_{k+2}}{f_{k+1}} </math>
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<math> x_{k+1}=\frac{s_{k+2}}{s_{k+1}} </math>
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Since that was what we wanted to show, we see that the terms in our continued fraction are represented by ratios of Fibonacci numbers.
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So by the principle of mathematical induction, we have shown that the terms in our continued fraction are represented by ratios of consecutive Fibonacci numbers.
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The exact continued fraction is <math> x_{\infty} = \lim_{n\rightarrow \infty}\frac{f_{n+1}}{f_n} =\varphi </math>.
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The exact continued fraction is
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:<math> x_{\infty} = \lim_{n\rightarrow \infty}\frac{s_{n+1}}{s_n} =\varphi </math>.
}}|NumChars=75}}
}}|NumChars=75}}
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==Proof of the Golden Ratio's Irrationality==
==Proof of the Golden Ratio's Irrationality==
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{{SwitchPreview|ShowMessage=Click to expand|HideMessage=Click to hide|PreviewText= |FullText=
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{{SwitchPreview|ShowMessage=Click to expand|hideMessage=Click to hide|PreviewText= |FullText=
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Remarkably, the Golden Ratio is irrational, despite the fact that we just proved that is approximated by a ratio of Fibonacci numbers.
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Remarkably, the Golden Ratio is <balloon title="A number is irrational if it cannot be expressed as the fraction between two integers.">irrational</balloon>, despite the fact that we just proved that is approximated by a ratio of Fibonacci numbers.
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We will use the method of <balloon title="A method of proving a statement true by assuming that it's false and showing this assumption would logically lead to a statement that is already known to be untrue.">contradiction</balloon> to prove that the golden ratio is <balloon title="A number is irrational if it cannot be expressed as the fraction between two integers.">irrational</balloon>.
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We will use the method of <balloon title="A method of proving a statement true by assuming that it's false and showing this assumption would logically lead to a statement that is already known to be untrue.">contradiction</balloon> to prove that the golden ratio is irrational.
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Suppose <math>\varphi </math> is rational. Then it can be written as fraction in lowest terms <math> \varphi = b/a</math>, where a and b are integers.
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Suppose <math>\varphi </math> is rational. Then it can be written as fraction in lowest terms <math> \varphi = b/a</math>, where ''a'' and ''b'' are integers.
Our goal is to find a different fraction that is equal to <math> \varphi </math> and is in lower terms. This will be our contradiction that will show that <math> \varphi </math> is irrational.
Our goal is to find a different fraction that is equal to <math> \varphi </math> and is in lower terms. This will be our contradiction that will show that <math> \varphi </math> is irrational.
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Now, since we know
Now, since we know
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<math> \frac{b}{a}=\frac{a+b}{b} </math>
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:<math> \frac{b}{a}=\frac{a+b}{b} </math>
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we see that <math> b^2=a(a+b) </math> by cross multiplication. Writing this all the way out gives us <math> b^2=a^2+ab </math>.
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we see that <math> b^2=a(a+b) </math> by cross multiplication. Foiling this expression gives us <math> b^2=a^2+ab </math>.
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Rearranging this gives us <math> b^2-ab=a^2 </math>, which is the same as <math> b(b-a)=a^2 </math>.
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Rearranging this gives us <math> b^2-ab=a^2 </math>, which is the same as :<math> b(b-a)=a^2 </math>.
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Dividing both sides of the equation by <math> (b-a) </math> and <math> a </math> gives us that
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Dividing both sides of the equation by ''a(b-a)'' gives us
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<math> \frac{b}{a}=\frac{a}{b-a} </math>.
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:<math> \frac{b}{a}=\frac{a}{b-a} </math>.
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Since <math> \varphi=\frac{b}{a} </math>, we can see that <math> \varphi=\frac{a}{b-a} </math>.
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Since <math> \varphi=\frac{b}{a} </math>, this means <math> \varphi=\frac{a}{b-a} </math>.
Since we have assumed that a and b are integers, we know that b-a must also be an integer. Furthermore, since <math> a<b </math>, we know that <math> \frac{a}{b-a} </math> must be in lower terms than <math> \frac{b}{a} </math>.
Since we have assumed that a and b are integers, we know that b-a must also be an integer. Furthermore, since <math> a<b </math>, we know that <math> \frac{a}{b-a} </math> must be in lower terms than <math> \frac{b}{a} </math>.
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:*Markowsky. “Misconceptions about the Golden Ratio.” College Mathematics Journal. Vol 23, No 1 (1992). pp 2-19.
:*Markowsky. “Misconceptions about the Golden Ratio.” College Mathematics Journal. Vol 23, No 1 (1992). pp 2-19.
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:*http://www.contracosta.edu/math/pentagrm.htm
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Revision as of 12:42, 18 July 2012


The Golden Ratio
Fields: Algebra and Geometry
Image Created By: azavez1
Website: The Math Forum

The Golden Ratio

The golden number, often denoted by lowercase Greek letter "phi", is
{\varphi}=\frac{1 + \sqrt{5}}{2} = 1.61803399....

The term golden ratio refers to any ratio which has the value phi. The image to the right illustrates dividing and subdividing a rectangle into the golden ratio. The result is fractal-like. This page explores real world applications for the golden ratio, common misconceptions about the golden ratio, and multiple derivations of the golden number.


Contents

Basic Description

Does the Mona Lisa exhibit the golden ratio?
Does the Mona Lisa exhibit the golden ratio?
The golden number, approximately 1.618, is called golden because many geometric figures involving this ratio are often said to possess special beauty. Be that true or not, the ratio has many beautiful and surprising mathematical properties. The Greeks were aware of the golden ratio, but did not consider it particularly significant with respect to aesthetics. It was not called the "divine" proportion until the 15th century, and was not called "golden" ratio until the 18th century. [1]
Markowsky has determined the above dimensions to be incorrect.
Markowsky has determined the above dimensions to be incorrect.

Since then, it has been claimed that the golden ratio is the most aesthetically pleasing ratio, and claimed that this ratio has appeared in architecture and art throughout history. Among the most common such claims are that the Parthenon and Leonardo Da Vinci's Mona Lisa use the golden ratio. Even more esoteric claims propose that the golden ratio can be found in the human facial structure, the behavior of the stock market, and the Great Pyramids.
However, such claims have been criticized in scholarly journals as wishful thinking or sloppy mathematical analysis. Additionally, there is no solid evidence that supports the claim that the golden rectangle is the most aesthetically pleasing rectangle.[2]


Misconceptions about the Golden Ratio

Many rumors and misconceptions surround the golden ratio. There have been many claims that the golden ratio appears in art and architecture. In reality, many of these claims involve warped images and large margins of error. One claim is that the Great Pyramids exhibit the golden ratio in their construction. This belief is illustrated below.


In his paper, Misconceptions about the Golden Ratio, George Markowsky disputes this claim, arguing that the dimensions assumed in the picture are not anywhere close to being correct. Another belief is that a series of golden rectangles appears in the Mona Lisa. However, the placing of the golden rectangles seems arbitrary. Markowsky also disputes the belief that the human body exhibits the golden ratio. To read more, click here!

What do you think?

George Markowsky argues that, like the Mona Lisa, the Parthenon does not exhibit a series of golden rectangles (discussed below). Do you think the Parthenon was designed with the golden ratio in mind or is the image below simply a stretch of the imagination?

[3]

A Geometric Representation

The Golden Ratio in a Line Segment

Image:Animation2.gif


The golden number can be defined using a line segment divided into two sections of lengths a and b. If a and b are appropriately chosen, the ratio of a to b is the same as the ratio of a + b to a and both ratios are equal to \varphi. The line segment above (left) exhibits the golden proportion. The line segments above (right) are also examples of the golden ratio. In each case,

\frac{{\color{Red}\mathrm{red}}+\color{Blue}\mathrm{blue}}{{\color{Blue}\mathrm{blue}} }= \frac{{\color{Blue}\mathrm{blue}} }{{\color{Red}\mathrm{red}} }= \varphi .

The Golden Rectangle

A golden rectangle is any rectangle where the ratio between the sides is equal to phi. When the sides lengths are proportioned in the golden ratio, the rectangle is said to possess the golden proportions. A golden rectangle has sides of length \varphi \times r and 1 \times r where r can be any constant. Remarkably, when a square with side length equal to the shorter side of the rectangle is cut off from one side of the golden rectangle, the remaining rectangle also exhibits the golden proportions. This continuing pattern is visible in the golden rectangle below.

Image:Coloredfinalrectangle1.jpg


Triangles

The golden number, \varphi, is used to construct the golden triangle, an isoceles triangle that has legs of length \varphi \times r and base length of 1 \times r where r can be any constant. It is above and to the left. Similarly, the golden gnomon has base \varphi \times r and legs of length 1 \times r. It is shown above and to the right. These triangles can be used to form regular pentagons (pictured above) and pentagrams.

The pentgram below, generated by the golden triangle and the golden gnomon, has many side lengths proportioned in the golden ratio.

Image:Star1.jpg
\frac{{\color{SkyBlue}\mathrm{blue}} }{{\color{Red}\mathrm{red}} } = \frac{{\color{Red}\mathrm{red}} }{{\color{Green}\mathrm{green}} } = \frac{{\color{Green}\mathrm{green}} }{{\color{Magenta}\mathrm{pink}} } = \varphi .

These triangles can be used to form fractals and are one of the only ways to tile a plane using pentagonal symmetry. Pentagonal symmetry is best explained through example. Below, we have two fractal examples of pentagonal symmetry. Images that exhibit pentagonal symmetry have five symmetry axes. This means that we can draw five lines from the image's center, and all resulting divisions are identical.


A More Mathematical Explanation

Note: understanding of this explanation requires: *Algebra, Geometry

An Algebraic Derivation of Phi

<span class="_togglegroup _toggle_initshow _toggle _toggler toggle- [...]

An Algebraic Derivation of Phi

How can we derive the value of \varphi from its characteristics as a ratio? We may algebraically solve for the ratio (\varphi) by observing that ratio satisfies the following property by definition:

\frac{b}{a} = \frac{a+b}{b} = \varphi

Let  r denote the ratio :

r=\frac{a}{b}=\frac{a+b}{a}.

So

r=\frac{a+b}{a}=1+\frac{b}{a} which can be rewritten as
1+\cfrac{1}{a/b}=1+\frac{1}{r} thus,
r=1+\frac{1}{r}

Multiplying both sides by r, we get

 {r}^2=r+1

which can be written as:

r^2 - r - 1 = 0 .

Applying the quadratic formula , we get r = \frac{1 \pm \sqrt{5}} {2}.

The ratio must be positive because we can not have negative line segments or side lengths. Because the ratio has to be a positive value,

r=\frac{1 + \sqrt{5}}{2} = 1.61803399... =\varphi.




Continued Fraction Representation and Fibonacci Sequences

The golden ratio can also be written as what is called a continued fraction,a fraction of infinite length whose denominator is a quantity plus a fraction, which latter fraction has a similar denominator, and so on. This is done by using recursion.


We have already solved for \varphi using the following equation:

{\varphi}^2-{\varphi}-1=0.

We can add one to both sides of the equation to get

{\varphi}^2-{\varphi}=1.

Factoring this gives

 \varphi(\varphi-1)=1 .

Dividing by \varphi gives us

\varphi -1= \cfrac{1}{\varphi }.

Adding 1 to both sides gives

\varphi =1+ \cfrac{1}{\varphi }.

Substitute in the entire right side of the equation for  \varphi in the bottom of the fraction.

\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{\varphi } }

Substituting in again,

\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\varphi}}}


\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\cdots}}}

Continuing this substitution forever, the last infinite form is a continued fraction

If we evaluate truncations of the continued fraction by evaluating only part of the continued fraction, replacing \varphi with 1, we produce the ratios between consecutive terms in the Fibonacci sequence.

\varphi \approx 1 + \cfrac{1}{1} = 2

\varphi \approx 1 + \cfrac{1}{1+\cfrac{1}{1}} = 3/2

\varphi \approx 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1} } } = 5/3

\varphi \approx 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1+\cfrac{1}{1}}}} = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{2}}} =1 + \cfrac{1}{1 + \cfrac{2}{3}} = 8/5

Thus we discover that the golden ratio is approximated in the Fibonacci sequence.

1,1,2,3,5,8,13,21,34,55,89,144...\,

1/1 = 1
2/1 = 2
3/2 = 1.5
8/5 = 1.6
13/8 = 1.625
21/13 = 1.61538462...
34/21 = 1.61904762...
55/34 = 1.61764706...
89/55 = 1.61818182...


\varphi = 1.61803399...\,

As you go farther along in the Fibonacci sequence, the ratio between the consecutive terms approaches the golden ratio. Many real-world applications of the golden ratio are related to the Fibonacci sequence. For more real-world applications of the golden ratio click here!

In fact, we can prove that the ratio between terms in the Fibonacci sequence approaches the golden ratio by using mathematical Induction.

Since we have already shown that

 \varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\cdots}}} ,

we only need to show that each of the terms in the continued fraction is the ratio of Fibonacci numbers as shown above.

First, let

 x_1=1,
 x_2=1+\frac{1}{1}=1+\frac{1}{x_1} ,
 x_3= 1+\frac{1}{1+\frac{1}{1}}=1+\frac{1}{x_2} and so on so that
 x_n=1+\frac{1}{x_{n-1}} .

These are just the same truncated terms as listed above. Let's also denote the terms of the Fibonacci sequence as

 s_n=s_{n-1}+s_{n-2} where s_1=1,s_2=1,s_3=2,s_4=3 etc.


We want to show that

 x_n=\frac{s_{n+1}}{s_n} for all n.

First, we establish our base case. We see that

 x_1=1=\frac{1}{1}=\frac{s_2}{s_1} , and so the relationship holds for the base case.

Now we assume that

 x_k=\frac{s_{k+1}}{s_{k}} for some  1 \leq k < n (This step is the inductive hypothesis). We will show that this implies that
 x_{k+1}=\frac{s_{(k+1)+1}}{s_{k+1}}=\frac{s_{k+2}}{s_{k+1}} .



By our assumptions about x1,x2...xn, we have

 x_{k+1}=1+\frac{1}{x_k} .

By our inductive hypothesis, this is equivalent to

x_{k+1}=1+\frac{1}{\frac{s_{k+1}}{s_{k}}}.

Now we only need to complete some simple algebra to see

 x_{k+1}=1+\frac{s_k}{s_{k+1}}
 x_{k+1}=\frac{s_{k+1}+s_k}{s_{k+1}}

Noting the definition of s_n=s_{n-1}+s_{n-2}, we see that we have

 x_{k+1}=\frac{s_{k+2}}{s_{k+1}}

So by the principle of mathematical induction, we have shown that the terms in our continued fraction are represented by ratios of consecutive Fibonacci numbers.

The exact continued fraction is

 x_{\infty} = \lim_{n\rightarrow \infty}\frac{s_{n+1}}{s_n} =\varphi .



Proof of the Golden Ratio's Irrationality


Remarkably, the Golden Ratio is irrational, despite the fact that we just proved that is approximated by a ratio of Fibonacci numbers. We will use the method of contradiction to prove that the golden ratio is irrational.

Suppose \varphi is rational. Then it can be written as fraction in lowest terms  \varphi = b/a, where a and b are integers.

Our goal is to find a different fraction that is equal to  \varphi and is in lower terms. This will be our contradiction that will show that  \varphi is irrational.

First note that the definition of  \varphi = \frac{b}{a}=\frac{a+b}{b} implies that  b > a since clearly  b+a>b and the two fractions must be equal.


Now, since we know

 \frac{b}{a}=\frac{a+b}{b}

we see that  b^2=a(a+b) by cross multiplication. Foiling this expression gives us  b^2=a^2+ab .

Rearranging this gives us  b^2-ab=a^2 , which is the same as : b(b-a)=a^2 .

Dividing both sides of the equation by a(b-a) gives us

 \frac{b}{a}=\frac{a}{b-a} .

Since  \varphi=\frac{b}{a} , this means  \varphi=\frac{a}{b-a} .

Since we have assumed that a and b are integers, we know that b-a must also be an integer. Furthermore, since  a<b , we know that  \frac{a}{b-a} must be in lower terms than  \frac{b}{a} .

Since we have found a fraction of integers that is equal to  \varphi , but is in lower terms than  \frac{b}{a} , we have a contradiction:  \frac{b}{a} cannot be a fraction of integers in lowest terms. Therefore  \varphi cannot be expressed as a fraction of integers and is irrational.



For More Information

  • Markowsky. “Misconceptions about the Golden Ratio.” College Mathematics Journal. Vol 23, No 1 (1992). pp 2-19.




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References

  1. "Golden ratio", Retrieved on 20 June 2012.
  2. "Misconceptions about the Golden Ratio", Retrieved on 24 June 2012.
  3. "Parthenon", Retrieved on 16 May 2012.

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